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Extract numbers from a string based on a another string [duplicate]

This question already has answers here:
Java Regex group 0
(2 answers)
Closed 4 years ago.
I am trying to write a regex for a string which has a format [digit] [to] [digit] eg. 1 to 5 in which if I find a word "to" from a given string i want to extract the number before and after, I have tried this and it's not working.
Pattern p = Pattern.compile("([0-9]+)\\bto\\b([0-9]+)");
Matcher m = p.matcher("1 to 5");
m.find();
System.out.println(m.group(0));
System.out.println(m.group(1));
System.out.println(m.group(2));
Expected o/p
1
to
5

Consider adding a group for the to part.
Also for the space, you want \\s not \\b:
Pattern p = Pattern.compile("([0-9]+)\\s(to)\\s([0-9]+)");
Matcher m = p.matcher("1 to 5");
m.find();
System.out.println(m.group(1));
System.out.println(m.group(2));
System.out.println(m.group(3));
And as said in the comments :
" Group zero denotes the entire pattern"

Is it necessary that you must use regex. If not, you can use String functions.
String s="23 to 34";
String toString="to";
if(s.contains(toString)){
int startIndex=s.indexOf(toString);
int endIndex=startIndex+(toString).length();
String s1=s.substring(0, startIndex); //get the first number
String s2=s.substring(endIndex); //get the second number
System.out.println(s1.trim()); // Removing any whitespaces
System.out.println(toString);
System.out.println(s2.trim();
}

Need a Regex that extracts a string between two "delimiting" strings [duplicate]

This question already has answers here:
Java Regex Capturing Groups
(4 answers)
Closed 6 years ago.
I need to get the string between by_ and _on.
So far I have this, but don't understand how to truncate the actual "string delimiters":
by_(.*)_on
Sample input:
Files_by_wesasegeaazedude_on_January_26.jpg
Current Match:
by_wesasegeaazedude_on
Needed Match:
wesasegeaazedude

Your expression is good*. All you need to do is extracting the content of the first capturing group:
Pattern regex = Pattern.compile("by_(.*)_on");
String str = "Files_by_wesasegeaazedude_on_January_26.jpg";
Matcher m = regex.matcher(str);
if (m.find()) {
String res = m.group(1);
}
Demo.
* Well, almost good. If you expect inputs with multiple file names on the same line, you may want to consider using reluctant qualifier, i.e. by_(.*?)_on

I would do this without regular expressions.
int start = str.indexOf("by_");
int end = str.indexOf("_on", start + 1); // or lastIndexOf("_on"), for greedy match.
assert start > 0 && end > start;
String part = str.substring(start + 3, end);

You can simply use positive lookarounds:
String regex = "(?<=by_).*(?=_on)";
What this regex does is:
match anything: .*
that is preceded by by_: (?<=by_)
and followed by _on: (?=_on)

regular expression to extract string from java code [duplicate]

This question already has answers here:
Regex to replace all string literals in a Java file
(4 answers)
Closed 8 years ago.
The sample source code to match is
String string="welcome";
String k="a\"welcome";
I am using "(\"[^(\")]*\")" regex in java.
But this extracts
0:"welcome"
0:"a\"
Expected output is
0:"welcome"
0:"a\"welcome"
What change should i make in regex to get the expected output ?
Java source :
private static String pattern1="(\"[^(\")]*\")";
public void getStrings(){
Pattern r = Pattern.compile(pattern1);
Matcher m = r.matcher("String string=\"welcome\";\n" +
"String k=\"a\\\"welcome\";");
while(m.find()){
System.out.println("0:"+m.group(0));
}
}

Just use lookahead and lookbehind in your regex,,
(?<==)(".*?")(?=;)
Get the value from group index 1.
DEMO
Pattern r = Pattern.compile("(?<==)(\".*?\")(?=;)");
Matcher m = r.matcher("String string=\"welcome\";\n" +
"String k=\"a\\\"welcome\";");
while(m.find()){
System.out.println("0:"+m.group(1));
}
Output:
0:"welcome"
0:"a\"welcome"
OR
Use the greediness of *,
Pattern r = Pattern.compile("(\".*\")");
OR
It skips the double quotes which are preceded by a backslash,
Pattern r = Pattern.compile("(\\\".*?(?<=[^\\\\])\\\")");

Why do you even bother with variable assignment. You know that everything within "" is a string.
"(.+)"\s*; should do it just fine.

String pattern matching with regular expression in java [duplicate]

This question already has answers here:
Java regular expressions and dollar sign
(5 answers)
Closed 8 years ago.
I'm trying pattern matching expression for a below string. But it doesn't work. could you anybody help me on this ? Only Alphanumeric and underscore allowed inside,Both side
$ sign will be there. Ex strings: Test_1,23_test_2,test3.
String text = "$test_1$";
Pattern p = Pattern.compile("$([A-Za-z0-9_])$");
Matcher m = p.matcher(text);
m.matches();
if (m.find()) {
System.out.println("Matched: " + m.group(1));
} else {
System.out.println("No match.");
}

Your regex should be:
Pattern p = Pattern.compile("(\\$[A-Za-z0-9_]*\\$)");

You could simply do...
s.matches("\\$[a-zA-Z0-9_]*\\$")

$ is a regex meta character and should be escaped, try this
Pattern p = Pattern.compile("\\$([A-Za-z0-9_]+)\\$");

Java Regular Expression checking for symbols [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Regular expression for excluding special characters
How would I check if a String contains a symbol? let's say I have this String
"SUGAR :::: SUGAR"
I would want to check if that string contains = the following Symbols
":,?,!##$%^&*()";
I tried this
Pattern p = Pattern.compile("[?,!,#,$,%,^,&,*,(,)]");
Matcher m = p.matcher("?");
boolean b = m.matches();
System.out.println(b);
But what if the text contains multiple occurrences of that symbol

Try this:
Pattern pat = Pattern.compile("[:?!##$%^&*()]");
String str = "SUGAR :::: SUGAR";
Matcher m = pat.matcher(str);
if (m.find()) {
System.out.println("string contains pattern");
}
The above will check if any part of the string contains at least one occurrence of any of the symbols in the pattern (no need to separate them with ,).

guava may be?
CharMatcher matcher = CharMatcher.anyOf(":,?,!##$%^&*()");
boolean result = matcher.matchesAnyOf("SUGAR :: SUGAR");
System.out.println(result);

Just simply match against a set containing these characters: [:,?!##$%^&*()]
If you want to check against the symbol -, position it to the end of the set: [...-]
There's no need to separate elements in a set by comma.
If you want to check for ANY non alphanumeric use \W instead: [\W_] (because \W does not match _)